If $T^k = Id$ for $k\ge 1$ then $T$ is diagonalizable

Question

Let $V$ a finite dimension space over $\mathbb{C}$ and $T:V\to V$, a linear transformation such that $T^k = Id$ for $k\ge 1$. Prove that $T$ is diagonalizable.

I'd be glad for an hint. How do I approach this question?

Thanks.

Answer 1

Hints:

1) The operator $\;T\;$ is regular (i.e., $\;\det T\neq 0\iff $ all its eigenvalues are non-zero)

2) Suppose $\;\lambda\neq 0\; $ , then over a field of characteristic $\;\neq 2\;$ (which contains all the eigenvalues of $\;T\;$ , say $\;\Bbb C\;$ )

$$\begin{pmatrix}\lambda&1&0\ldots&0\\ 0&\lambda&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots\end{pmatrix}^2=\begin{pmatrix}\lambda^2&2\lambda&0\ldots&0\\ 0&\ldots&\ldots&\ldots\\ \ldots&\ldots&\ldots&\ldots\end{pmatrix}$$

3) Deduce that $\;T\;$ has to be diagonalizable